Your output has a mistake, since it's giving -2 to left pixels. Let's add some coordinates to your image:
You start in raster scan order and hit step (1)(a): it's an outer border, so $NBD\leftarrow2$, $(i,j)\leftarrow(1,1)$, $(i_2,j_2)\leftarrow(1,0)$.
Let's call $P_{start}\equiv(i,j)$, and $P_{from}\equiv(i_2,j_2)$.
Then you move to step (2) and set its parent to 1.
Now the counterclockwise border following algorithm:
(3.1 aka InitPhase): search clockwise around $P_{start}$ starting from $P_{from}$ a nonzero pixel. You check in order: $(1,0)$, $(0,0)$, $(0,1)$, $(0,2)$, and eventually you find $(1,2)$. Now you set $(i_1,j_1)\leftarrow(1,2)$. Let's call $P_{start\mbox{-}from}\equiv(i_1,j_1)$. In our case we have the following result:
$P_{start}$ is yellow and $P_{start\mbox{-}from}$ is green. You need the starting point and the original direction to detect multiple passes through it, and avoid stopping too early.
(3.2 aka Setup): $(i_2,j_2)\leftarrow(1,2)$ and $(i_3,j_3)\leftarrow(1,1)$. Let's call $P_{cur}\equiv(i_3,j_3)$ and $P_{prev}\equiv(i_2,j_2)$.
(3.3 aka FindNext): search counterclockwise around $P_{cur}$ starting from the next of $P_{prev}$ a nonzero pixel. You check in order: $(0,2)$, $(0,1)$, $(0,0)$, $(1,0)$, $(2,0)$, and eventually you find $(2,1)$. Now you set $(i_4,j_4)\leftarrow(2,1)$. Let's call $P_{next}\equiv(i_4,j_4)$. In our case we have the following result:
$P_{prev}$ is pink, $P_{cur}$ is green, and $P_{next}$ is blue.
(3.4 aka SetValue): the image at $(i_3,j_3+1)=(1,2)$, the pixel to the right of $P_{cur}$, and at $P_{cur}$ is nonzero, so we are in case (b): $f(P_{cur})\leftarrow2$.
(3.5 aka MoveNext): $P_{prev} \leftarrow P_{cur}$ and $P_{cur} \leftarrow P_{next}$.
Let's do it again!
(3.3 aka FindNext): search counterclockwise around $P_{cur}$ starting from the next of $P_{prev}$ a nonzero pixel. You check in order: $(1,0)$, $(2,0)$, $(3,0)$ and eventually $P_{next}\leftarrow(3,1)$. We have the following result:
Watch out: here is the trick
(3.4 aka SetValue): the image at $(i_3,j_3+1)=(2,2)$, the pixel to the right of $P_{cur}$, is not a 0-pixels examined in the substep (3.3). In fact we examined $(1,0)$, $(2,0)$, $(3,0)$. So we are again in case (b): $f(P_{cur})\leftarrow2$.
Now you can keep going. The final result (before resuming the raster scan) will be:
Wow. This was so long to explain...
Here is the specific point in the article:
